Problem: In the diagram, $QRS$ is a straight line. What is the measure of $\angle RPS,$ in degrees? [asy]
pair Q=(0,0);
pair R=(1.3,0);
pair SS=(2.3,0);
pair P=(.8,1);

draw(P--Q--R--SS--P--R);

label("$Q$",Q,S);
label("$R$",R,S);
label("$S$",SS,S);
label("$P$",P,N);

label("$48^\circ$",Q+(.12,.05),NE);
label("$67^\circ$",P-(.02,.15),S);
label("$38^\circ$",SS+(-.32,.05),NW);

[/asy]
Answer: Since the sum of the angles in a triangle is $180^\circ,$ then \begin{align*}
\angle QPS &= 180^\circ - \angle PQS - \angle PSQ \\
&= 180^\circ - 48^\circ - 38^\circ \\
&= 94^\circ.
\end{align*}Therefore, \begin{align*}
\angle RPS &= \angle QPS - \angle QPR \\
&= 94^\circ - 67^\circ \\
&= \boxed{27^\circ}.
\end{align*}